Question: Solve for $x$ : $x^2 - 14x + 48 = 0$
The coefficient on the $x$ term is $-14$ and the constant term is $48$ , so we need to find two numbers that add up to $-14$ and multiply to $48$ The two numbers $-8$ and $-6$ satisfy both conditions: $ {-8} + {-6} = {-14} $ $ {-8} \times {-6} = {48} $ $(x {-8}) (x {-6}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -8) (x -6) = 0$ $x - 8 = 0$ or $x - 6 = 0$ Thus, $x = 8$ and $x = 6$ are the solutions.